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2 years ago

5

If a triangular cross-section with coordinates at (1, 1), (1, 4), and (3, 1) is rotated about the line x = 1, what will be the r

esulting three-dimensional object?
Cone
Cylinder
Sphere
Pyramid

2 answers:

love history [14]2 years ago

8 0

Answer:

COOOONNNNEEE!!!!

Mkey [24]2 years ago

4 0

Answer:

answer = cone

Step-by-step explanation:

just took the test and got it right! :)

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3 years ago

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2 years ago

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3 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1

charle [14.2K]

<u>Answer-</u> Length of the curve of intersection is 13.5191 sq.units

<u>Solution-</u>

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ $\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = 2 \cos t$

$\frac{dz}{dt} = \sin t-2cos t$

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

$= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}$

$=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}$

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$

Now evaluating the integral using calculator,

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$ $= 13.5191$

8 0

3 years ago